Theorem bnj98 30937

Description: Technical lemma for bnj150 30946. This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)

Assertion

Ref	Expression
bnj98	⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Proof of Theorem bnj98

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 3203	. . . . . 6 ⊢ 𝑖 ∈ V
2	1	sucid 5804	. . . . 5 ⊢ 𝑖 ∈ suc 𝑖
3	2	n0ii 3922	. . . 4 ⊢ ¬ suc 𝑖 = ∅
4		df-suc 5729	. . . . . 6 ⊢ suc 𝑖 = (𝑖 ∪ {𝑖})
5		df-un 3579	. . . . . 6 ⊢ (𝑖 ∪ {𝑖}) = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
6	4, 5	eqtri 2644	. . . . 5 ⊢ suc 𝑖 = {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})}
7		df1o2 7572	. . . . . . 7 ⊢ 1𝑜 = {∅}
8	6, 7	eleq12i 2694	. . . . . 6 ⊢ (suc 𝑖 ∈ 1𝑜 ↔ {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅})
9		elsni 4194	. . . . . 6 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} ∈ {∅} → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
10	8, 9	sylbi 207	. . . . 5 ⊢ (suc 𝑖 ∈ 1𝑜 → {𝑥 ∣ (𝑥 ∈ 𝑖 ∨ 𝑥 ∈ {𝑖})} = ∅)
11	6, 10	syl5eq 2668	. . . 4 ⊢ (suc 𝑖 ∈ 1𝑜 → suc 𝑖 = ∅)
12	3, 11	mto 188	. . 3 ⊢ ¬ suc 𝑖 ∈ 1𝑜
13	12	pm2.21i 116	. 2 ⊢ (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))
14	13	rgenw 2924	1 ⊢ ∀𝑖 ∈ ω (suc 𝑖 ∈ 1𝑜 → (𝐹‘suc 𝑖) = ∪ 𝑦 ∈ (𝐹‘𝑖) pred(𝑦, 𝐴, 𝑅))

Syntax hints:   → wi 4   ∨ wo 383   = wceq 1483   ∈ wcel 1990  {cab 2608  ∀wral 2912   ∪ cun 3572  ∅c0 3915  {csn 4177  ∪ ciun 4520  suc csuc 5725  ‘cfv 5888  ωcom 7065  1_𝑜c1o 7553   predc-bnj14 30754

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-v 3202  df-dif 3577  df-un 3579  df-nul 3916  df-sn 4178  df-suc 5729  df-1o 7560

This theorem is referenced by:  bnj150  30946